I want to differentiate the following equation by taking $\log$ with respect to $\theta$.
$\log (\theta^{ a_H+\alpha-1}(1-\theta)^{a_T+\beta-1})$
and have the result of the differentiation as below:
$\theta=\dfrac{a_H+\alpha-1}{a_H+\alpha+a_T+\beta-2}$
I am trying to learn it but not really good at doing the steps of algebra. Step by step calculations would be appreciated.
Let's assume:
$a=\alpha_H+\alpha-1$
and
$b=\alpha_T+\beta-1$
Therefore,
$\log (\theta^{ a_H+\alpha-1}(1-\theta)^{a_T+\beta-1})=\log\theta^a(1-\theta)^b=\log\theta^a+\log(1-\theta)^b={a\log\theta+b\log(1-\theta)}$
The derivative of $\log\theta$ with respect to $\theta$ is $\frac{1}{\theta}$, and the derivative of $\log(1-\theta)$ with respect to $\theta$ is $\frac{-1}{1-\theta}$ therefore, the derivative of the above equation is:
$\frac{a}{\theta}-\frac{b}{1-\theta}$
Apparently, you want the above derivative to become zero:
$\frac{a}{\theta}=\frac{b}{1-\theta}$
Solving the above equation is straightforward and results in:
$\theta=\frac{a}{a+b}$
which is what you want.
